A. Operator Relations A.1 Theorem 1 Let A and B be two non-commuting operators, then [1] α2 exp(αA)B exp(−αA)=B + α [A, B]+ [A, [A, B]] + .... (A.1) 2! Proof. Let f1(α)=exp(αA)B exp(−αA) , (A.2) then, one can expand f1 in Taylor series about the origin. We first evaluate the derivatives − − f1(α)=exp(αA)(AB BA)exp( αA) , so f1(0) = [A, B] . (A.3) Similarly − − f1 (α)=exp(αA)(A [A, B] [A, B] A)exp( αA) , so that f1 (0) = [A, [A, B]] . (A.4) Now, we write the Taylor’s expansion α2 f (α)=f (0) + αf (0) + f (0) + ... (A.5) 1 1 1 2! 1 or α2 exp(αA)B exp(−αA)=B + α [A, B]+ [A, [A, B]] + .. (A.6) 2! A particular case is when [A, B]=c,wherec is a c-number, then exp(αA)B exp(−αA)=B + αc , (A.7) in which case exp(αA) acts as a displacement operator. 364 A. Operator Relations A.2 Theorem 2: The Baker–Campbell–Haussdorf Relation Let A and B be two non-commuting operators such that [A, [A, B]] = [B,[A, B]] = 0 , (A.8) then α2 exp [α(A + B)] = exp αA exp αB exp − [A, B] (A.9) 2 α2 =expαB exp αA exp [A, B] . 2 Proof. Define f2(α) ≡ exp αA exp αB . (A.10) Then df (α) 2 =[A +exp(αA)B exp −(αA)]f (α) (A.11) dα 2 =(A + B + α [A, B])f2(α) , where in the last step, we used (A.6). Also, from the definition of f2(α), we can write df (α) 2 =exp(αA)A exp αB +exp(αA)exp(αB)B (A.12) dα =expαA exp αB [exp(−αB)A exp αB + B] = f2(α)(A + B + α [A, B]) . By comparing (A.11) and (A.12), we can see that f2(α)commuteswith (A+B +α [A, B]), thus one can integrate as a c-number differential equation, getting α2 α2 f (α)=exp (A + B)α + [A, B] =expα(A+B)exp [A, B] , (A.13) 2 2 2 thus obtaining the desired result. Another application of the Theorem 1 is taking A = aa†, (A.14) B = a or a† . As [n, a]=−a (A.15) A.3 Theorem 3: Similarity Transformation 365 and the higher order commutators also give a with alternating signs, thus α2 exp(αn)a exp(−αn)=a − αa + a + ... =exp(−α)a. (A.16) 2 Similarly exp(αn)a† exp(−αn)=exp(α)a† . (A.17) A.3 Theorem 3: Similarity Transformation exp(αA)f(B)exp−(αA)=f(exp(αA)(B)exp−(αA)) . (A.18) Proof. We start with the following identity [exp(αA)(B)exp−(αA)]n =exp(αA)B exp(−αA)exp(αA)B exp(−αA)... =exp(αA)Bn exp(−αA) . Then, for any function f(B) that can be expanded in power series, the Theorem 3 follows. As an interesting application, let us calculate exp −αa† + α∗a f(a, a†)exp αa† − α∗a = f[exp −αa† + α∗a a exp αa† − α∗a , exp −αa† + α∗a a† exp αa† − α∗a ] = f(a + α, a† + α∗) . Also exp −αa† f(a, a†)exp αa† = f(a + α, a†) , (A.19) exp (α∗a) f(a, a†)exp(−α∗a)=f(a, a† + α∗) , (A.20) exp (αn) f(a, a†)exp(−αn)=f[a exp(−α),a† exp(α)] (A.21) Other properties: One can easily show that da†l a, a†l = la†l−1 = , (A.22) da† dal a†,al = −lal−1 = − . da A more general version of the above relations is for a function f(a, a) which may be expanded in power series of a and a† ∂f(a, a†) a, f(a, a†) = , (A.23) ∂a† ∂f(a, a†) a†,f(a, a†) = − . (A.24) ∂a 366 A. Operator Relations Reference 1. Louisell, W.H.: Quantum Statistical Properties of Radiation. John Wiley, New York (1973) B. The Method of Characteristics We have a first-order partial differential equation: [1] Pp+ Qq = R, (B.1) where P = P (x, y, z),Q= Q(x, y, z),R= R(x, y, z), and ∂z ∂z p ≡ ,q ≡ , (B.2) ∂x ∂y and we wish to find a solution of (B.1), of the form z = f(x, y) . (B.3) The general solution of (B.1) is F (u, v)=0, (B.4) where F is an arbitrary function, and u(x, y, z)=c1 , (B.5) v(x, y, z)=c2 , is a solution of the equations dx dy dz = = . (B.6) P Q R Proof. If (B.5) are solutions of (B.6), then the equations ∂u ∂u ∂u dx + dy + dz =0, (B.7) ∂x ∂y ∂z and dx dy dz = = , P Q R must be compatible; thus, we must have 368 B. The Method of Characteristics Pux + Quy + Ruz =0, (B.8) and similarly for v Pvx + Qvy + Rvz =0. (B.9) On the other hand, if x and y are independent variables and z = z(x,y), then from (B.5), we get ∂z u + u =0, (B.10) x z ∂x ∂z u + u =0, y z ∂y and substituting (B.10) into (B.8) we get ∂z ∂z ∂u −P − Q + R =0, ∂x ∂y ∂z and (B.1) is satisfied. The second part of the proof is to show that the general solution of (B.1) is F (u, v)=0. (B.11) From (B.11), one writes ∂F ∂F ∂u ∂u ∂z ∂F ∂v ∂v ∂z = + + + =0, (B.12) ∂x ∂u ∂x ∂z ∂x ∂v ∂x ∂z ∂x ∂F ∂F ∂u ∂u ∂z ∂F ∂v ∂v ∂z = + + + =0. (B.13) ∂y ∂u ∂y ∂z ∂y ∂v ∂y ∂z ∂y We finally notice that (B.13) is satisfied considering (B.10). Example. Find the general solution of the equation ∂z ∂z x2 + y2 =(x + y)z. (B.14) ∂x ∂y In this case P = x2 , (B.15) Q = y2 , R =(x + y)z, and we have to find the solution of dx dy dz = = . (B.16) x2 y2 (x + y)z Reference 369 Integrating, first dx dy = , x2 y2 we get −1 −1 x + y = c1 . (B.17) On the other hand, x2 − dx − dy ( 2 1)dy dy dz = y = = , x2 − y2 x2 − y2 y2 (x + y)z from where we get x − y = c = v. (B.18) z 2 Combining (B.17) and (B.18), we get xy = c = u, (B.19) z 1 so the general solution can be put as xy x − y F , =0, (B.20) z z or if we write (B.20) in the equivalent form u = g(v) , (B.21) then the solution is xy x − y = g . (B.22) z z Reference 1. Sneddon, I.: Elements of Partial Differential Equations. (Mc-Graw Hill, New York (1957) C. Proof In this Appendix, we show the equation ⎡ ⎤ ⎣ − − − 2⎦ 2 − − δ(t tj)δ(t tk) S R ρaa = pRδ(t t ) . (C.1) j= k For regular pumping, one can put tj = t0 + jτ, where τ is the constant time interval between the atoms and t0 some arbitrary time origin [1]. In this case, there are no pumping fluctuations, and therefore, there are no correlations between the products of delta functions, that is δ(t − tj)δ(t − tk)S = δ(t − tj)S δ(t − tk)S (C.2) j,k j k = R2 . Now, we split the l.h.s. of the above equation in two parts 2 δ(t − tj)δ(t − tk)S + δ(t − tj)δ(t − tk)S = R , (C.3) j=k j=k 2 δ(t − tj)δ(t − tk)S + Rδ(t − t )=R , j= k thus proving the relation ⎡ ⎤ ⎣ − − − 2⎦ 2 − − δ(t tj)δ(t tk) S R ρaa = pRδ(t t ) j= k for p =1. In the Poissonian case, tj is totally uncorrelated from tk(j = k),so 2 δ(t − tj)δ(t − tk)S = δ(t − tj)S δ(t − tk)S = R , (C.4) j= k j k which proves (C.1) for p =0. Notice that in the above result, we are missing an atom in the second summation, so the above result is approximate, the approximation being very good when R>>1. (The error is of the order of R compared to R2.) A more general proof is found in the reference [2]. 372 C. Proof References 1. An excellent discussion on this point, as well and on noise supression in quantum optical systems is found in: Davidovich, L.: Rev. Mod. Phys., 68, 127 (1996) 2. Benkert, C., Scully, M.O., Bergou, J., Davidovich, L., Hillery, M., Orszag, M.: Phys. Rev. A, 41, 2756 (1990) D. Stochastic Processes in a Nutshell D.1 Introduction Classical Mechanics gives a deterministic view of the dynamical variables of a system. This of course is true, when one is not in a chaotic regime. On the other hand, in many cases, the system under study is only de- scribed by the time evolution of probability distributions. To show these ideas with an example, we take a look at the random walk in one dimension, by now, a classical problem [3]. A person moves in a line, taking random steps forward or backward, with equal probability, at fixed time intervals τ. Calling the position xn = na, then the probability that it occupies the site xn at time t is P (xn | t) and obeys the equation 1 1 P (x | t + τ)= P (x − | t)+ P (x | t) .